Problem 204 Solution

Question

On May 23, 2009, a craps player held the dice for 154 rolls at the Borgata casino in Atlantic City (source). What is the probability of going 154 rolls or longer in craps? Please give an expression of the answer and/or a numeric answer.

Answer

Solution

There are four possible states the shooter can be in. Let's define them as follows.

State 1 = Come out roll
State 2 = Point of 4 or 10
State 3 = Point of 5 or 9
State 4 = Point of 6 or 8

The probabilities are recursive. Let p(x,r) represent the probability of being in state x before roll r. Based on simple dice probabilities, p(x,r) can be expressed as follows:

(1) P(1,r) = (12/36)×p(1,r-1) + (3/36)×p(2,r-1) + (4/36)×p(3,r-1) + (5/36)× p(4,r-1)
(2) P(2,r) = (6/36)×p(1,r-1) + (27/36)×p(2,r-1)
(3) P(3,r) = (8/36)×p(1,r-1) + (26/36)×p(3,r-1)
(4) P(4,r) = (10/36)×p(1,r-1) + (25/36)×p(4,r-1)

In more plain simple English, equation (1) is saying that the probability of being in a come out roll before roll r is the sum of the following:

1. Product of the probability of being in a come out roll the previous turn, and the probability of staying in a come out roll (12/36).
2. Product of the probability of rolling for a 4 or 10 the previous roll, and the probability of making the point (3/36), resulting in a come out roll.
3. Product of the probability of rolling for a 5 or 9 the previous roll, and the probability of making the point (4/36), resulting in a come out roll.
4. Product of the probability of rolling for a 6 or 8 the previous roll, and the probability of making the point (5/36), resulting in a come out roll.

Equation (2) is saying the probability of rolling for a 4 or 10 before roll r is the sum of:

1. Product of the probability of being in a come out roll the previous turn, and the probability of rolling a 4 or 10 (6/36).
2. Product of the probability of rolling for a point of 4 or 10 the previous turn, and the probability of not rolling the desired point or a 7 (27/36).

Equation (3) is saying the probability of rolling for a 5 or 9 before roll r is the sum of:

1. Product of the probability of being in a come out roll the previous turn, and the probability of rolling a 5 or 9 (8/36).
2. Product of the probability of rolling for a point of 5 or 9 the previous turn, and the probability of not rolling the desired point or a 7 (26/36).

Equation (4) is saying the probability of rolling for a 6 or 8 before roll r is the sum of:

1. Product of the probability of being in a come out roll the previous turn, and the probability of rolling a 6 or 8 (10/36).
2. Product of the probability of rolling for a point of 6 or 8 the previous turn, and the probability of not rolling the desired point or a 7 (25/36).

Next, let's express the these four equations in the form of a matrix.

[ P(1,r) ]             [12   3   4   5]   [ P(1,r-1) ]
[ P(2,r) ]             [ 6  27   0   0]   [ P(2,r-1) ]
[ P(3,r) ] = (1/36) ×  [ 8   0  26   0] × [ P(3,r-1) ]
[ P(4,r) ]             [10   0   0  25]   [ P(4,r-1) ]

The intial roll is always a come out roll, so the initial state is:

[ 1 ]
[ 0 ]
[ 0 ]
[ 0 ]

The probability of each state before the 154th roll is expressed as follows.  The reason we take the matrix to the 153rd power, and not the 154th, is there are 153 transformations since the initial state.

[ P(1,154) ]                 [12   3   4   5]^153 [1] 
[ P(2,154) ]                 [ 6  27   0   0]     [0]
[ P(3,154) ] = (1/36)^153 ×  [ 8   0  26   0] ×   [0]
[ P(4,154) ]                 [10   0   0  25]     [0]
     

The exact answer is: 

P(1,154) = 3.12763 × 10-11
P(2,154) = 4.63460 × 10-11
P(3,154) = 4.95558 × 10-11
P(4,154) = 5.17044 × 10-11

The sum of these probabilities is 1.78882 × 10-10 = 1 in 5,590,264,072